This is very much not about the order in which you roll dice or whether you roll them at the same time or one after the other. It is about whether you examine the conditional probability or the overall probability.
Given die one shows a 6, die two has a 6-probability of $1/6$.
Not given that any one of them has a 6, the probability of them both getting 6 is $1/6\times1/6=1/36$
TL;DR: if $p$ = 1/6 and you want to know how large $n$ needs to be 98% sure the dice is fair (to within 2%), $n$ needs to be at least $n$ ≥ 766.
Let $n$ be the number of rolls and $X$ the number of rolls that land on some specified side. Then $X$ follows a Binomial(n,p) distribution where $p$ is the probability of getting that specified side.
By the central limit theorem, we know that
$$\sqrt{n} (X/n - p) \to N(0,p(1-p))$$
Since $X/n$ is the sample mean of $n$ Bernoulli$(p)$ random variables. Hence for large $n$, confidence intervals for $p$ can be constructed as
$$\frac{X}{n} \pm Z \sqrt{\frac{p(1-p)}{n}}$$
Since $p$ is unknown, we can replace it with the sample average $\hat{p} = X/n$, and by various convergence theorems, we know the resulting confidence interval will be asymptotically valid. So we get confidence intervals of the form
$$\hat{p} \pm Z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$
with $\hat{p} = X/n$. I'm going to assume you know what $Z$-scores are. For example, if you want a 95% confidence interval, you take $Z=1.96$. So for a given confidence level $\alpha$ we have
$$\hat{p} \pm Z_\alpha \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$
Now let's say you want this confidence interval to be of length less than $C_\alpha$, and want to know how big a sample we need to make this case. Well this is equivelant to asking what $n_\alpha$ satisfies
$$Z_\alpha \sqrt{\frac{\hat{p}(1-\hat{p})}{n_\alpha}} \leq \frac{C_\alpha}{2}$$
Which is then solved to obtain
$$n_\alpha \geq \left(\frac{2 Z_\alpha}{C_\alpha}\right)^2 \hat{p}(1-\hat{p})$$
So plug in your values for $Z_\alpha$, $C_\alpha$, and estimated $\hat{p}$ to obtain an estimate for $n_\alpha$. Note that since $p$ is unknown this is only an estimate, but asymptotically (as $n$ gets larger) it should be accurate.
Best Answer
As other people have pointed out in comments, the correct answer to the question "what is the probability of rolling another 6 given that I have rolled a 6 prior to it?" is indeed $\frac{1}{6}$. This is because the die rolls are assumed (very reasonably so) to be independent of each other. This means that past rolls of the die does not affect future die rolls.
Expressed mathematically, independence of two variables $X$ and $Y$ imply that $Pr(Y=y | X = x) = Pr(Y = y)$.
Letting $X$ be a variable denoting the outcome of the first die roll and $Y$ be a variable for the second die roll, we can use the definition of independence to arrive to the conclusion that $Pr(Y=6 | X = 6) = Pr(Y = 6)=1/6$.
The reason that the answer is not 1/36 is due to the fact that we are making a conditional statement. We are saying "given that we already have rolled a six in the first roll". This means that we are not interested in the likelihood of that first roll occuring. We are only interested in what happens next.
It might be helpful to enumerate all possible outcomes here. I have done this below in the form {x, y}, where x is the outcome in the first roll and y in the second.
{1, 1} {1, 2} {1, 3} {1, 4} {1, 5} {1, 6}
{2, 1} {2, 2} {2, 3} {2, 4} {2, 5} {2, 6}
{3, 1} {3, 2} {3, 3} {3, 4} {3, 5} {3, 6}
{4, 1} {4, 2} {4, 3} {4, 4} {4, 5} {4, 6}
{5, 1} {5, 2} {5, 3} {5, 4} {5, 5} {5, 6}
{6, 1} {6, 2} {6, 3} {6, 4} {6, 5} {6, 6}
Now, the probability you are interested in is the event {6, 6}. If you give the information that you are in the last row (which corresponds to having rolled a 6 in the first roll), you only have six possibilities of outcomes. Only one of them is a "success", so the probability of that event is 1/6.
Edit:
After re-reading the OP's question, it appears that I have missed part of the question. The question there seems to be regarding the following scenario:
The question is there: What is the probability that this procedure results in two sixes having been rolled? Equivalently: What is the probability that this procedure results in us rolling a six in step 2?
The answer to this question is indeed 1/36. Heuristically, the reason for this is that we now are not conditioning on something that has happened anymore. We are instead asking for the probability of an event that can occur after we go through a procedure.
Let us now prove that the probability is 1/36. Letting once again $X$ be the result of the first roll and $Y$ the result of the second roll. We are interested in $Pr(Y=6)$. Note that if $X\neq 6$ then the probability that $Y=6$ is zero since the second die won't be rolled. Thus $Pr(Y=6\mid X\neq6)=0$. We use the law of total probability to note that $Pr(Y=6)=\underset{x=1}{\overset{6}{\sum}}Pr(Y=6 \mid X=x) \cdot Pr(X=x)$.
Now since $Pr(Y=6 \mid X=x)=0$ $\forall x\neq 6$, we see that
$Pr(Y=6) = 0+0+0+0+0+Pr(Y=6\mid X=6)\cdot Pr(X=6)$.
This simplifies to $Pr(Y=6) = \frac{1}{6}\cdot\frac{1}{6}=\frac{1}{36}$ which completes the proof.